Prove by induction that \(\displaystyle \L 1\, + \,\frac{1}{2^2 }\, + \,\frac{1}{3^2}\, +\, ...\, + \,\frac{1}{n^2 }\, \,2\, - \,\frac{1}{n}\) for all positive integers of n.
Solving for P(1), we get the LS = RS, which makes it true, we then assume it works for P(k). We then must prove it works for P(K+1)
Trying to work it out, I came up with:
\(\displaystyle \L 1 + \;\frac{1}{2^2 }\; + \;\frac{1}{3^2 }\; + \;...\; + \;\frac{1}{n^2}\; + \;\frac{1}{\left( {k + 1} \right)^2 } \; \;2\; - \;\frac{1}
{k}\; + \;\frac{1}{\left( {k + 1} \right)^2 }\)
\(\displaystyle \L 1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; 2\; - \;\frac{{\left( {k + 1} \right)^2 }}
{{k\left( {k + 1} \right)^2 }}\; + \;\frac{k}
{{k\left( {k + 1} \right)^2 }}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k + 1} \right)^2 + k}}
{{k\left( {k + 1} \right)^2 }}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k^2 + 2k + 1} \right) + k}}
{{k\left( {k^2 + 2k + 1} \right)}}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k^2 + k + 1} \right)}}
{{k\left( {k^2 + 2k + 1} \right)}}\)
I think the answer I am trying to derive is: \(\displaystyle \[
2 - \;\frac{1}
{{k + 1}}
\]\)
I am stuck here. I was wondering if somebody could let me know if what I am doing is right, if I made an error somewhere, and could help me out so I can complete the question.
Thanks.
Solving for P(1), we get the LS = RS, which makes it true, we then assume it works for P(k). We then must prove it works for P(K+1)
Trying to work it out, I came up with:
\(\displaystyle \L 1 + \;\frac{1}{2^2 }\; + \;\frac{1}{3^2 }\; + \;...\; + \;\frac{1}{n^2}\; + \;\frac{1}{\left( {k + 1} \right)^2 } \; \;2\; - \;\frac{1}
{k}\; + \;\frac{1}{\left( {k + 1} \right)^2 }\)
\(\displaystyle \L 1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; 2\; - \;\frac{{\left( {k + 1} \right)^2 }}
{{k\left( {k + 1} \right)^2 }}\; + \;\frac{k}
{{k\left( {k + 1} \right)^2 }}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k + 1} \right)^2 + k}}
{{k\left( {k + 1} \right)^2 }}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k^2 + 2k + 1} \right) + k}}
{{k\left( {k^2 + 2k + 1} \right)}}\)
\(\displaystyle \L
1 + \;\frac{1}
{{2^2 }}\; + \;\frac{1}
{{3^2 }}\; + \;...\; + \;\frac{1}
{{n^2 }}\; + \;\frac{1}
{{\left( {k + 1} \right)^2 }}\; \;2\; - \;\frac{{\left( {k^2 + k + 1} \right)}}
{{k\left( {k^2 + 2k + 1} \right)}}\)
I think the answer I am trying to derive is: \(\displaystyle \[
2 - \;\frac{1}
{{k + 1}}
\]\)
I am stuck here. I was wondering if somebody could let me know if what I am doing is right, if I made an error somewhere, and could help me out so I can complete the question.
Thanks.